# What is the vertex form of the equation of the parabola with a focus at (-3,-9) and a directrix of y=-10?

${\left(x - - 3\right)}^{2} = 2 \left(y - - \frac{19}{2}\right)$

#### Explanation:

The vertex of a parabola is always in between the focus and the directrix

From the given, the directrix is lower than the focus. Therefore the parabola opens upward.

p is 1/2 of the distance from the directrix to the focus

$p = \frac{1}{2} \left(- 9 - - 10\right) = \frac{1}{2} \cdot 1 = \frac{1}{2}$

vertex #(h, k)=(-3, (-9+(-10))/2)=(-3, -19/2)

${\left(x - h\right)}^{2} = 4 p \left(y - k\right)$

${\left(x - - 3\right)}^{2} = 4 \cdot \left(\frac{1}{2}\right) \left(y - - \frac{19}{2}\right)$

${\left(x - - 3\right)}^{2} = 2 \left(y - - \frac{19}{2}\right)$

see the graph with directrix $y = - 10$

graph{((x--3)^2-2(y--19/2))(y+10)=0[-25,25,-13,13]}

have a nice day from the Philippines